Chapter 10 - Section 10.4 - Reactions with Gaseous Reactants and Products - Checkpoint - Page 445: 10.4.1

$\therefore\frac{1}{2}$ mole of Na will give 11.4 L of $Cl_2$

$2Na_{(s)} + Cl_2(g) = 2NaCl_{(s)}$ From the above equation 2 mole of Na gives 1 mole of $Cl_2$ At STP volume of 1 mole of $Cl_2$ is 22.4 L Now, 1 mole of Na will give $\frac{1}{2}$ mole of $Cl_2$ At STP volume of 1/2 mole of $Cl_2$ is 11.2 L